Quiz 3 - Acids and Bases | Acids and Bases

The conjugate acid is formed by adding a proton (H⁺) to the base. For HPO₄²⁻​, the conjugate acid is formed by adding one proton, which results in H₂PO₄⁻​. This species is the conjugate acid of HPO₄^2−​.

Step 1: Calculate the moles of NaOH and HCl

• Moles of NaOH: n_NaOH = [NaOH] x V_NaOH = 0.0657 M x 0.200 L = 0.01314 moles

• Moles of HCl: n_HCl = [HCl] x V_HCl = 0.107 M x 0.140 L = 0.01498 moles

Step 2: Determine the limiting reagent

• NaOH neutralizes HCl in a 1:1 ratio.
• HCl: 0.01498 moles > NaOH: 0.01314 moles

Thus, NaOH is the limiting reagent, and all of it will react with HCl.

Step 3: Calculate the moles of excess HCl after the reaction
After neutralization: Excess HCl moles = 0.01498 moles − 0.01314 moles = 0.00184 moles

Step 4: Determine the concentration of H⁺ ions
The total volume of the solution is: 200 mL + 140 mL + 160 mL = 500 mL = 0.500L
Concentration of H⁺ ions (from excess HCl): [H⁺] = $\frac{0.00184 \mathrm{moles}}{0.500 \mathrm{L}}$ ​= 0.00368 M

Step 5: Calculate the pH
pH= − log⁡ [H⁺] = − log ⁡(0.00368) ≈ 2.43

Step 1: Write the equation for the hydrolysis of NaOCl

OCl⁻ + H₂O ⇌ HOCl + OH⁻

The equilibrium expression for this reaction is based on the base dissociation constant K_b​.

Step 2: Calculate the K_b​ from K_a​

We are given the K_a​ of HOCl: K_a = 2.8 x 10⁻⁸

The relationship between K_a​ and K_b​ for a conjugate acid-base pair is: K_b = $\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}}$

where K_w​ is the ionization constant for water (1.0 x 10⁻¹⁴ at 25°C). Thus: K_b = $\frac{1.0 \mathrm{x} {10}^{-14}}{2.8 \mathrm{x} {10}^{-8}}$ = 3.57 x 10^-7

Step 3: Set up the ICE table and use the K_b​ expression

Let x represent the concentration of OH⁻ at equilibrium.

K_b = $\frac{\left[\mathrm{HOCl}\right]\left[{\mathrm{HO}}^{-}\right]}{\left[{\mathrm{OCl}}^{-}\right]}$ = $\frac{{\mathrm{x}}^2}{0.65}$

Step 4: Solve for x

x² = (3.57 x 10⁻⁷) x 0.65 ≈ 2.32 x 10⁻⁷ ⇒ x ≈ 4.8 x 10⁻⁴ M

Thus, the concentration of OH⁻ is approximately 4.8 x 10⁻⁴ M

Step 1: Write the expression for the ionization of formic acid

The ionization of formic acid can be represented by the equation: HCOOH ⇌ H⁺ + HCOO⁻

Step 2: Set up the ICE table and expression for K_a

Let x represent the concentration of H⁺ and HCOO⁻ ions at equilibrium. The initial concentration of HCOOH is 0.10 M, and it decreases by x upon ionization.

K_a = $\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{HCOO}}^{-}\right]}{\left[\mathrm{HCOOH}\right]}$ = $\frac{{\mathit{x}}^{\mathit{2}}}{\mathit{0}\mathit{.}\mathit{10}\mathit{ }\mathit{-}\mathit{ }\mathit{x}}$ =1.8 x 10^-4

Since x is expected to be small, we approximate 0.10 − x ≈ 0.10, simplifying the equation:

1.8 x 10⁻⁴ = $\frac{x^2}{0.10}$ ⇒ x ≈ 4.24 x 10⁻³ M

Step 3: Calculate the percentage ionization

The percentage ionization is given by:

Percentage ionization = $\frac{\left[H^{+}\right]}{{\left[HCOOH\right]}_{initial}}$ x 100 = $\frac{4.24 \mathrm{x} {10}^{-3}}{0.10}$ x 100 ≈ 4.2%

• The reverse reaction is: HCO₂⁻ + H₂O ⇌ HCO₂H + OH⁻
  The equilibrium constant for this reverse reaction is K_b​, which is related to the K_a​ of formic acid and the ionization constant of water, K_w​.

• The relationship between K_a, K_b​, and K_w​ is: Kw = K_a x K_b
  where K_w = 1.0 x 10^-14.
  Therefore, K_b = $\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}}$ = $\frac{1.0 \mathrm{x} {10}^{-14}}{1.8 \mathrm{x} {10}^{-4}}$ = 5.6 x 10^-11

• The equilibrium constant for the reaction given in the question is the reciprocal of K_b​, because the reaction is the reverse of the base dissociation reaction. Thus:
  K = $\frac{1}{{\mathrm{K}}_{\mathrm{b}}}$ = $\frac{1}{5.6 \mathrm{x} {10}^{-11}}$ = 1.8 x 10¹⁰

HI (hydroiodic acid) is a strong acid, which means it fully dissociates in water. Therefore, the concentration of H⁺ ions in solution will be equal to the concentration of the acid itself.

Since HI fully dissociates: [H⁺] = 0.043 M

pH = − log ⁡[H⁺] = −log ⁡(0.043) ≈ 1.37

Step 1: Calculate the K_a from the given data

The percent ionization of a weak acid is related to the concentration of the acid as follows:
Percent ionization = 3.8% = $\frac{\left[{\mathrm{H}}^{+}\right]}{{\left[\mathrm{HA}\right]}_0}$ x 100 with [HA]₀ = 0.10 M

Let x be the concentration of H⁺ ions. Thus, from the percent ionization formula:

[H⁺] = x = $\frac{3.8}{100}$ x 0.10 = 3.8 x 10^-3 M

Now, let's calculate K_a for glycolic acid:

K_a = $\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]}$ = $\frac{{\mathrm{x}}^2}{0.10 - \mathrm{x}}$ = 1.444 x 10^-4

Step 2: Calculate [H⁺] in the 0.50M solution

Now, for the 0.50M solution, we can use the same K_a​ value calculated from the first step:

K_a = $\frac{{\left[{\mathrm{H}}^{+}\right]}^2}{0.50}$ ⇒ [H⁺] = $\sqrt{0.50 {\mathrm{K}}_{\mathrm{a}}}$ = $\sqrt{7.22 \times {10}^{-5}}$ = 8.5 x 10^-3 M

Step 3: Calculate the % ionization in the 0.50M solution

The percent ionization in the 0.50 M solution is:

Percent Ionization = $\frac{8.5 \times {10}^{-3}}{0.50}$ x 100 = 1.7%

Step 1: Calculate the [OH⁻]

Given that the pH is 11.50, we can calculate the pOH first:

pOH = 14 − pH = 14 − 11.50 = 2.50

Next, calculate the concentration of hydroxide ions: [OH⁻] = 10^−pOH = 10^−2.50 ≈ 3.16 x 10⁻³ M

Step 2: Use the relationship between K_b​ and K_a

We are given the pK_a of NH₄⁺​, which is 9.24. Using the relationship between K_a and K_b​:

K_b = $\frac{{\mathrm{K}}_{\mathrm{w}}}{{\mathrm{K}}_{\mathrm{a}}}$ = $\frac{1.0 \mathrm{x} {10}^{-14}}{{10}^{-9.24}}$ = 5.75 x 10^-5

Step 3: Set up the equilibrium expression for ammonia

The reaction for ammonia in water is: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻

The expression for the base dissociation constant K_b​ is: K_b =​ $\frac{\left[{{\mathrm{NH}}_4}^{+}\right]\left[{\mathrm{OH}}^{-}\right]}{\left[{\mathrm{NH}}_3\right]}$

Assuming that the concentration of [NH₄⁺] is approximately equal to [OH⁻] at equilibrium (since it comes from the same ionization reaction), we can substitute:

K_b = $\frac{{\left[{\mathrm{OH}}^{-}\right]}^2}{\left[{\mathrm{NH}}_3\right]}$ ⇒ [NH₃] = $\frac{{\left[{\mathrm{OH}}^{-}\right]}^2}{{\mathrm{K}}_{\mathrm{b}}}$ = $\frac{{(3.16 \mathrm{x} {10}^{-3})}^2}{5.75 \mathrm{x} {10}^{-5}}$ ≈ 0.58 M